Question

$\frac{(1+cot\theta +tan\theta )(sin\theta -cos\theta )}{se{c}^3\theta -cose{c}^3\theta }=si{n}^2\theta co{s}^2\theta$

Answer 1

$LHS=\frac{(1+cot\theta +tan\theta )(sin\theta -cos\theta )}{se{c}^3\theta -cose{c}^3\theta }$

$=\frac{1+\frac{cos\theta }{sin\theta }+\frac{sin\theta }{cos\theta }}{\frac{1}{co{s}^3\theta }-\frac{1}{si{n}^3\theta }}$ $\left[\begin{array}{c}\because cot\theta =\frac{cos\theta }{sin\theta },tan\theta =\frac{sin\theta }{cos\theta }\\ sec\theta =\frac{1}{cos\theta },cosec\theta =\frac{1}{sin\theta }\end{array}\right]$

$=\frac{1+\frac{co{s}^2\theta +si{n}^2\theta }{sin\theta cos\theta }}{\frac{si{n}^3\theta -co{s}^3\theta }{co{s}^3\theta si{n}^3\theta }}(sin\theta -cos\theta )$

$=\frac{(sin\theta cos\theta +1)si{n}^3\theta co{s}^3\theta }{sin\theta cos\theta .(si{n}^3\theta -co{s}^3\theta )}(sin\theta -cos\theta )(\because si{n}^2\theta +co{s}^2\theta =1)$

$=\frac{(1+sin\theta cos\theta )si{n}^2\theta co{s}^2\theta .(sin\theta -cos\theta )}{(sin\theta -cos\theta )(si{n}^2+co{s}^2\theta +sin\theta cos\theta )}(\because a^3-b^2=(a-b\left)\right(a^2+b^2+ab)$

$=\frac{(1+sin\theta cos\theta ).si{n}^2\theta co{s}^2\theta }{(1+sin\theta cos\theta )}=si{n}^2\theta co{s}^2\theta =RHS$

Hence proved.